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Hausdorff Space |
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| Jan23-09, 09:56 PM | #1 |
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Hausdorff Space
1. The problem statement, all variables and given/known data
Show that X is Hausdorff if and only if the diagonal [tex]\Delta = \{x \times x | x \in X \}[/tex] is closed in [tex]X \times X[/tex]. 2. Relevant equations Definition of Hausdorff Space (T2) : A topological space in which distinct points have disjoint neighborhoods. 3. The attempt at a solution |
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| Jan24-09, 12:18 AM | #2 |
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If you haven't already, express the closedness of [tex]\Delta[/tex] in [tex]X\times X[/tex] (which I assume has the product topology) in terms of the openness of its complement. Now have a go at proving each direction (neither is more difficult than the other) if you haven't already, and please show us your efforts. |
| Jan24-09, 02:49 AM | #3 |
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If X is Hausdorff, the diagonal [tex] \Delta [/tex] is closed in [tex]X\times X[/tex]. Assume X is Hausdorff. Now, we have two distinct points x, y and disjoint open sets U, V containing x, y, respectively. The basis element [tex]U \times V[/tex] containing [tex](x,y) \in X \times X[/tex] should not intersect [tex] \Delta [/tex] by the assumption given to the Hausdorff property. For every [tex] (x,y) \notin \Delta [/tex], we have a basis element in [tex] X \times X [/tex] containing (x,y), which does not intersect [tex] \Delta [/tex]. Thus, [tex]X \times X \setminus \Delta is open[/tex] and we conclude [tex] \Delta [/tex] is a closed set in [tex]X \times X[/tex] . <- If the diagonal [tex] \Delta [/tex] is closed in [tex]X\times X[/tex], X is Haudorff. Supppose [tex] \Delta [/tex] is closed in [tex]X\times X[/tex]. Then, [tex]X \times X \setminus \Delta [/tex] is open. Let [tex](x,y) \in X \times X[/tex] and [tex] x \neq y [/tex]. For [tex](x,y) \notin \Delta[/tex], we have a basis element [tex] U \times V [/tex] in [tex]X \times X [/tex] containing (x, y). We remain to show U and V are disjoint. Suppose on the contrary that U and V are not disjoint. Then, there is an element [tex] (z,z) \in X times X [/tex] which belongs to both U and V. Contradicting the fact that x and y are distinct. Thus, X is Hausdorff. |
| Jan24-09, 02:31 PM | #4 |
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Hausdorff SpaceNow how 'bout you tackle your other question without copying down the solution from an external source. It's the only way you'll learn topology. |
| Jan24-09, 05:19 PM | #5 |
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Can you please give me a link of an external source you mentioned?
It's my self study of topology (I am not even majoring in math) and I don't need to copy the external source to show you something to impress you. Rather, I just did for my self study purpose and asked an advice if someone finds an error in my attempt to the solution. |
| Jan24-09, 08:00 PM | #6 |
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